Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{-4x + 40}{-3x^2 - 18x} \div \dfrac{x - 9}{x^2 - 3x - 54} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4x + 40}{-3x^2 - 18x} \times \dfrac{x^2 - 3x - 54}{x - 9} $ First factor the quadratic. $q = \dfrac{-4x + 40}{-3x^2 - 18x} \times \dfrac{(x + 6)(x - 9)}{x - 9} $ Then factor out any other terms. $q = \dfrac{-4(x - 10)}{-3x(x + 6)} \times \dfrac{(x + 6)(x - 9)}{x - 9} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -4(x - 10) \times (x + 6)(x - 9) } { -3x(x + 6) \times (x - 9) } $ $q = \dfrac{ -4(x - 10)(x + 6)(x - 9)}{ -3x(x + 6)(x - 9)} $ Notice that $(x - 9)$ and $(x + 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -4(x - 10)\cancel{(x + 6)}(x - 9)}{ -3x\cancel{(x + 6)}(x - 9)} $ We are dividing by $x + 6$ , so $x + 6 \neq 0$ Therefore, $x \neq -6$ $q = \dfrac{ -4(x - 10)\cancel{(x + 6)}\cancel{(x - 9)}}{ -3x\cancel{(x + 6)}\cancel{(x - 9)}} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $q = \dfrac{-4(x - 10)}{-3x} $ $q = \dfrac{4(x - 10)}{3x} ; \space x \neq -6 ; \space x \neq 9 $